Log and exp problems.

Question

The question, $$\ln 4 - 2\ln(x+1) = \ln e$$ Solve for $x$

Here is what I did: $$\ln \frac {4}{(x+1)^2} = \ln e$$ Hence, $$\frac {4}{(x+1)^2} = e$$ Unable to move further.

Please note my textbook advances 2 possible answers $-3$ and $1$

Answer 1

Divide by $e$ and multiply by $(x+1)^2$. Then expand the square:

$$x^2 + 2x + 1 = \frac{4}{e}$$

Hence

$$x^2 + 2x + 1 - \frac{4}{e} = 0$$

Second degree equation:

$$x_{1,\ 2} = \frac{-2\pm \sqrt{4 - 4\cdot\left(1 - \frac{4}{e}\right)}}{2} = \frac{-2 \pm \sqrt{4 - 4 + \frac{16}{e}}}{2} = \frac{-2\pm \sqrt{\frac{16}{e}}}{2} = \frac{-2\pm 4\sqrt{e^{-1}}}{2} = $$ $$x_{1,\ 2} = -1\pm 2\sqrt{\frac{1}{e}}$$

Of Course

One of the solutions is not valid since you need $(x+1)^2 > 0$

Answer 2

From $\frac{4}{(x+1)^2} = e$, you can solve for $x$ normally:

$$\frac{4}{e} = (x+1)^2$$

$$\pm \frac{2}{\sqrt{e}} = x+1$$

$$x = \pm \frac{2}{\sqrt{e}}-1$$

We see that $x = \frac{2}{\sqrt{e}}-1$ since the other solution would be invalid for $ln(x+1)$.